A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 11.5 s.

(a) How tall is the tower?
_________ m

(b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s2, what is the period there?______S

a. h = g*T^2/4pi^2.

h = 9.8*(11.5)^2/39.4 = 32.9 m.

b. T^2 = 4pi^2*(L/g).
T^2 = 39.4*(32.9/1.67) = 776.2
T = 27.9 s. = Period

Pi meaning

Pi (π) is a mathematical constant that represents the ratio of the circumference of a circle to its diameter. It is approximately equal to 3.14159, but is an irrational number with decimal digits that go on infinitely without repeating in a pattern. Pi is used in many mathematical and scientific calculations involving circles, spheres, and other curved shapes.

Give examples of distractive interference wave

(a) How tall is the tower?

Oh, it's so tall that even giraffes get dizzy looking up! Alright, let's calculate. The period of a pendulum is given by the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. We're given that the period is 11.5 s, so let's plug in the values. However, we need the acceleration due to gravity in order to calculate the height of the tower. So, let's assume a regular Earth setting and use g = 9.8 m/s²:

11.5 s = 2π√(L/9.8 m/s²)

To find the height of the tower (L), we need some math magic:

√(L/9.8 m/s²) = 11.5 s / (2π)

Now, let's square both sides to get rid of that square root:

L/9.8 m/s² = (11.5 s / (2π))²

L/9.8 m/s² = 18.53 s² / π²

L = (18.53 s² / π²) * 9.8 m/s²

L = 18.53 * 9.8 m² / π²

After doing the calculations, we find that the height of the tower is approximately 287.1 meters.

(b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s², what is the period there?

Well, well, well, our little pendulum is going on a lunar adventure! The formula for the period of a pendulum on the Moon can be derived the same way as on Earth but with our new gravity value:

T = 2π√(L/g_moon)

We're given the original period T (11.5 s), so we can substitute the values and solve for L:

11.5 s = 2π√(L/1.67 m/s²_moon)

Squaring both sides:

√(L/1.67 m/s²_moon) = 11.5 s / (2π)

L/1.67 m/s²_moon = (11.5 s / (2π))²

L/1.67 m/s²_moon = 18.53 s² / π²

Solving for L:

L = (18.53 s² / π²) * 1.67 m/s²_moon

L = 18.53 * 1.67 m²_moon / π²

After crunching those numbers, we find that the period of the pendulum on the Moon would be approximately 2.25 seconds. Have a pendulous time there!

To find the height of the tower, we can use the formula for the period of a pendulum, which is given by:

T = 2π * √(L/g)

Where:
T is the period of the pendulum
L is the length of the pendulum
g is the acceleration due to gravity

In this case, we are given the period of the pendulum (T = 11.5 s). We need to find the height of the tower (L) and the acceleration due to gravity (g).

(a) To find the height of the tower:
1. Rearrange the formula to solve for L:

L = (T/2π)^2 * g

2. Substitute the values into the formula:

L = (11.5 s / (2π))^2 * g

3. Since we don't have the value of g yet, we can assume the standard value for the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2.

L = (11.5 s / (2 * 3.14))^2 * 9.8 m/s^2

Calculating this equation will give us the height of the tower in meters.

(b) To find the period of the pendulum on the Moon:

1. We are given that the free-fall acceleration on the Moon is 1.67 m/s^2. We need to find the new period (T') of the pendulum.

T' = 2π * √(L/g')

Where:
T' is the new period of the pendulum on the Moon
L is the length of the pendulum (height of the tower)
g' is the acceleration due to gravity on the Moon (1.67 m/s^2)

2. Rearrange the formula to solve for T':

T' = 2π * √(L/g')

3. Substitute the values into the formula:

T' = 2π * √(L/1.67 m/s^2)

Calculating this equation will give us the period of the pendulum on the Moon in seconds (S).